Trig: prove equal

G

Guest

Guest
My math book shows how to do questions like this by working both sides until they're equal, but my math teacher wants us to work the left side until it is equal to the right.

I can do most of them. I've just got myself stuck on this one.

tanxsin^(2)x + tanxcos^(2)x = tanx

So we're supposed to take the right side and manipulate it using 8 basic identities to make it equal to tanx. I tried changing tanx to sinx/cosx, but that didn't seem to work. Help?

**Edit.. I meant left to right. Oops.

galactus

Super Moderator
Staff member
Solve it from left to right and then work backward.

$$\displaystyle tan(x)sin^{2}(x)+tan(x)cos^{2}(x)=tan(x)$$

$$\displaystyle tan(x)(sin^{2}(x)+cos^{2}(x))=tan(x)$$

$$\displaystyle tan(x)(1)=tan(x)$$

G

Guest

Guest
galactus said:
Solve it from left to right and then work backward.

$$\displaystyle tan(x)sin^{2}(x)+tan(x)cos^{2}(x)=tan(x)$$

$$\displaystyle tan(x)(sin^{2}(x)+cos^{2}(x))=tan(x)$$

$$\displaystyle tan(x)(1)=tan(x)$$
I don't understand how you got from
$$\displaystyle tan(x)sin^{2}(x)+tan(x)cos^{2}(x)=tan(x)$$
to
$$\displaystyle tan(x)(sin^{2}(x)+cos^{2}(x))=tan(x)$$, though.

Is is an $$\displaystyle ab^{2} + ac^{2} = a(b^{2}+c^{2})$$ thing?
Because it completely never occurred to me that you could do that, or that it was true.

**edit.. my first time with that kind of coding so of course I messed it up a bit.

galactus

Super Moderator
Staff member
I just factored out tan(x) and knew that $$\displaystyle sin^{2}(x)+cos^{2}(x)=1$$

Yes, it's that abc thing and you can do that. Your coding looks great for a newbie.

G

Guest

Guest
Okay, great. It never even occurred to me that I could do it that way.

And I just copied your coding from when I quoted you, haha. It takes me forever to do though.

Thanks a lot!

Oh, while I'm here, I have another question. I did a question for an assignment and I'm pretty sure it's right, but I'd like to have it checked.

Allison noted that competitive swimmers do a series of dolphin kicks upon diving into the water. She thought that the legs moved sinusoidally. The independent variable is time and dependent is the height above and below the water. At what times are the swimmer's legs 5 cm above their resting position?

time(s)|height(cm)
0.7|0
0.9|15
1.1|0
1.3|-15
1.5|0
etc, I'm sure you could figure the rest out.

I figured out that the period would be 0.8 seconds, meaning there was a horizontal stretch of 1/450.

The formula I came up with:

$$\displaystyle (1/15)y=sin450(x+0.1)$$

I then substituted y for 5 and came out with:
$$\displaystyle 450(x+0.1)=19.47+360K$$
$$\displaystyle 450(x+0.1)=160.53 +360K$$

$$\displaystyle x=0.0567+0.8K[br]$$
$$\displaystyle x=0.2567+0.8K[br]$$
$$\displaystyle KEI$$